Question: Divide the following complex numbers. $ \dfrac{-2-6i}{-1-3i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-1+3i}$ $ \dfrac{-2-6i}{-1-3i} = \dfrac{-2-6i}{-1-3i} \cdot \dfrac{{-1+3i}}{{-1+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-2-6i) \cdot (-1+3i)} {(-1-3i) \cdot (-1+3i)} = \dfrac{(-2-6i) \cdot (-1+3i)} {(-1)^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-2-6i) \cdot (-1+3i)} {(-1)^2 - (-3i)^2} = $ $ \dfrac{(-2-6i) \cdot (-1+3i)} {1 + 9} = $ $ \dfrac{(-2-6i) \cdot (-1+3i)} {10} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-2-6i}) \cdot ({-1+3i})} {10} = $ $ \dfrac{{-2} \cdot {(-1)} + {-6} \cdot {(-1) i} + {-2} \cdot {3 i} + {-6} \cdot {3 i^2}} {10} $ Evaluate each product of two numbers. $ \dfrac{2 + 6i - 6i - 18 i^2} {10} $ Finally, simplify the fraction. $ \dfrac{2 + 6i - 6i + 18} {10} = \dfrac{20 + 0i} {10} = 2 $